Question 200633
You are looking for the time, t, when h(t) = 0, so...
{{{h(t) = -16t^2+48t+160}}} Substitute h(t) = 0.
{{{0 = -16t^2+48t+160}}}  Rewrite in standard form:
{{{-16t^2+48t+160 = 0}}} First factor -16.
{{{-16(t^2-3t-10) = 0}}} Notice the change of sign in each term in the parentheses as a result of factoring the -16. Now apply the zero product rule.
{{{t^2-3t-10 = 0}}} Solve by factoring.
{{{(t+2)(t-5) = 0}}} Apply the zero product rule.
{{{(t+2) = 0}}} or {{{(t-5) = 0}}} so then...
{{{t = -2}}} or {{{t = 5}}} Discard the negative solution as the time, t, should be a positive value.
{{{highlight(t = 5)}}}seconds.